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Differential Equations for Applications

by: Mary Veum

Differential Equations for Applications MATH 3410

Marketplace > University of Connecticut > Mathematics (M) > MATH 3410 > Differential Equations for Applications
Mary Veum
GPA 3.97


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This 5 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 3410 at University of Connecticut taught by Staff in Fall. Since its upload, it has received 7 views. For similar materials see /class/205835/math-3410-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15
Math 3410 Fall 2009 Second semester differential equations 1 Some review 11 First order equations We will need to know how to do separable equations and linear equations A separable equation is one like dy 2 7 t dt 9 We can rewrite this as d y 1 7 7 tdt Integrating both sides7 ii it C y 2 so 1 y 7 1 39 it c A linear equation is one like 2 it 9 It i One multiplies by an integrating factor pgf 521ntt2 to get tzy Qty R 01 tsz t3 which leads to 2 714 tyilt 0 andthen ylt2 4 13239 When the linear equation has constant coef cients and is homogeneous ie7 the right hand side is 07 things are much easier To solve y 7 4y 0 we guess a solutions of the form y 6 so y requot Then re 7 4e 07 or r 4 and therefore the solution is y 06 To identify 0 one needs an initial condition7 eg7 y0 2 Then so we then have For non hornogeneous equations7 such as 17 4y 63 one way to solve it is to solve the homogeneous equation y 7 4y 07 and then use y 664 yp where yp is a particular solution One way to nd a particular solution is to make an educated guess If we guess yp A65 then we have y 7 4 3A63t 7 4A 7 and this will equal 63 if A 71 We conclude the solution to the non hornogeneous equation is y 664 7 63 12 Series From calculus we have the Taylor series 2 3 em1x m 2 4 cosz17 gi and 3 5 s1nx7 7 lf 239 V71 substituting and doing some algebra shows that e cos z39 sinx 2 Second order linear 21 Applications First consider a spring hung from the ceiling with a weight hanging from it Let u be the distance the weight is below equilibrium There is a restoring force upwards of amount ku by Hooke7s law There is darnping resistance against the motion which is iRu And the net force is related to accelera tion by Newton7s laws so ku 7 RM F mu This leads to mu Ru 7 ku 0 If there is an external force acting on the spring then the right hand side is replaced by The second example is that of a circuit with a resistor inductance coil and capacitor hooked up in series Let I be the current Q the charge R the resistance L the inductance and C the capacitance We know that I dQdt The voltage drop across the resistor is IR across the capacitor QC397 and across the inductance coil L So if Et is the potential put into the current7 l Et LQ RQ 5Q Sometimes this is differentiated to give 1 E t LI RI 51 22 Linear constant coef cients homogeneous Let7s look at an example y 7 5y 4y 0 From Math 2117 we know a way of solving this Let 1 y and this one equation becomes a system 711 5117419 We then set up matrices7 where and the equation is X 7 AX We assume W w1 w2 and that our solution is of the form X We for some 7 wl and LU2 We will review this method later when we want to generalize it7 but lets look at an easier method 4


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